The area of the triangle formed by the points $(2, 3, 5), (- 1, 3, 2)$ and $(3, 5, - 2)$ is

18 \sqrt{2}

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9

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18

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9 \sqrt{2}

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Solution

The correct option is D $9 \sqrt{2}$
Assume points $A, B, C$ of $△ A B C$ be $A (2, 3, 5), B (- 1, 3, 2)$ and $C (3, 5. - 2)$
$\to A B : 3^i + 3^k$
$\to A C :^i + 2^j - 7^k$
$\to A B \times \to A C = (3^i + 3^k) \times (^i + 2^j - 7^k)$
Area of $△ A B C = \frac{| \to A B \times \to A C |}{2} = 9 \sqrt{2}$

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If area of the triangle formed by the lines $y^{2} - 9 x y + 18 x^{2} = 0$ and y=9 is A sq. unit find the value of 4A.

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